Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of these two vector spaces of cardinalities $c$ and $c\times c = c$, so they are isomorphic as vector spaces over $\mathbb{Q}$.

Is there a way to prove that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups which does not use AC? Are there models of set theory in which these groups are not isomorphic? I'm also curious whether there is a proof (perhaps using AC) which does not make use of this vector space machinery, though I'm fairly sure the above proof is the simplest.